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Proof step 5 from proof lastn-restn-01 (previous step)

# x = n0 + n1 + 1, ∀ x, n0. # x = n0 + n1 → lastn(n1, x) = restn(n0, x) ⊦ lastn(n1 + 1, x) = restn(n0, x)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 6:
⊦ lastn(n + 1, x) = lastn(n, x.butlast) + x.last
Proof step 7:
# x = n0 + n1 + 1, ∀ x, n0. # x = n0 + n1 → lastn(n1, x) = restn(n0, x) ⊦ lastn(n1, x.butlast) + x.last = restn(n0, x)

Rule argument

spec: inst: name: lastN-rec seq: ⊦ lastn(n + 1, x) = lastn(n, x.butlast) + x.last subst: [x, n] → [x, n1] flags: rewrite, current

Additional information

Used terms: x, n1